std::set<Key,Compare,Allocator>::operator= - cppreference.com
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(1) | |
| (2) | ||
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(since C++11) (until C++17) |
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(since C++17) | |
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(3) | (since C++11) |
Replaces the contents of the container.
1) Copy assignment operator. Replaces the contents with a copy of the contents of other.
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If |
(since C++11) |
2) Move assignment operator. Replaces the contents with those of other using move semantics (i.e. the data in other is moved from other into this container). other is in a valid but unspecified state afterwards.
If std::allocator_traits<allocator_type>::propagate_on_container_move_assignment::value is true, the allocator of *this is replaced by a copy of that of other. If it is false and the allocators of *this and other do not compare equal, *this cannot take ownership of the memory owned by other and must move-assign each element individually, allocating additional memory using its own allocator as needed. In any case, all elements originally belonging to *this are either destroyed or replaced by element-wise move-assignment.
3) Replaces the contents with those identified by initializer list ilist.
Parameters
| other | - | another container to use as data source |
| ilist | - | initializer list to use as data source |
Return value
*this
Complexity
1) Linear in the size of *this and other.
2) Linear in the size of *this unless the allocators do not compare equal and do not propagate, in which case linear in the size of *this and other.
3) O(NlogN) in general, where N is size() + ilist.size(). Linear if ilist is sorted with respect to value_comp().
Exceptions
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1-3) May throw implementation-defined exceptions. |
(until C++17) |
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1,3) May throw implementation-defined exceptions. 2) noexcept
specification:
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(since C++17) |
Notes
After container move assignment (overload (2)), unless element-wise move assignment is forced by incompatible allocators, references, pointers, and iterators (other than the end iterator) to other remain valid, but refer to elements that are now in *this. The current standard makes this guarantee via the blanket statement in [container.reqmts]/67, and a more direct guarantee is under consideration via LWG issue 2321.
Example
The following code uses operator= to assign one std::set to another:
#include <initializer_list> #include <iostream> #include <iterator> #include <set> void print(auto const comment, auto const& container) { auto size = std::size(container); std::cout << comment << "{ "; for (auto const& element : container) std::cout << element << (--size ? ", " : " "); std::cout << "}\n"; } int main() { std::set<int> x{1, 2, 3}, y, z; const auto w = {4, 5, 6, 7}; std::cout << "Initially:\n"; print("x = ", x); print("y = ", y); print("z = ", z); std::cout << "Copy assignment copies data from x to y:\n"; y = x; print("x = ", x); print("y = ", y); std::cout << "Move assignment moves data from x to z, modifying both x and z:\n"; z = std::move(x); print("x = ", x); print("z = ", z); std::cout << "Assignment of initializer_list w to z:\n"; z = w; print("w = ", w); print("z = ", z); }
Output:
Initially:
x = { 1, 2, 3 }
y = { }
z = { }
Copy assignment copies data from x to y:
x = { 1, 2, 3 }
y = { 1, 2, 3 }
Move assignment moves data from x to z, modifying both x and z:
x = { }
z = { 1, 2, 3 }
Assignment of initializer_list w to z:
w = { 4, 5, 6, 7 }
z = { 4, 5, 6, 7 }